\(2,\\ a,=x\left(x^3-1\right)=x\left(x-1\right)\left(x^2+x+1\right)\\ b,=\left(x-2y\right)\left(x+2y\right)-2\left(x-2y\right)=\left(x-2y\right)\left(x+2y-2\right)\\ c,=4x^2-y^2+2y-1=4x^2-\left(y-1\right)^2=\left(2x-y+1\right)\left(2x+y-1\right)\\ d,=3x^2+3x-2x-2=3x\left(x+1\right)-2\left(x+1\right)=\left(3x-2\right)\left(x+1\right)\\ 3,\\ a,\Rightarrow x^2-x-2=x^2\\ \Rightarrow x=-2\\ b,\Rightarrow x\left(x-3\right)-4\left(x-3\right)=0\\ \Rightarrow\left(x-4\right)\left(x-3\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=4\end{matrix}\right.\\ c,\Rightarrow x\left(2x-1\right)^2=0\Rightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{2}\end{matrix}\right.\\ d,\Rightarrow x^3-x^2+4x^2-4x+7x-7=0\\ \Rightarrow\left(x-1\right)\left(x^2+4x+7\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=1\\\left(x+2\right)^2+3=0\left(vô.lí\right)\end{matrix}\right.\\ \Rightarrow x=1\)
3)a,⇒x2−x−2=x2⇒x=−2b,⇒x(x−3)−4(x−3)=0⇒(x−4)(x−3)=0⇒[x=3x=4c,⇒x(2x−1)2=0⇒⎡⎣x=0x=12d,⇒x3−x2+4x2−4x+7x−7=0⇒(x−1)(x2+4x+7)=0⇒[x=1(x+2)2+3=0(vô.lí)⇒x=1
2b)=(x−2y)(x+2y)−2(x−2y)=(x−2y)(x+2y−2)
mik giúp r nhé
